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a) Ta có: \(3-\left(17-x\right)=-12\)
\(\Leftrightarrow3-17+x+12=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)
\(\Leftrightarrow\left|x-9\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{24;-6\right\}\)
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\(1)\frac{1}{5}+\frac{2}{11}< \frac{x}{55}< \frac{2}{5}+\frac{1}{55}\)
\(\Rightarrow\frac{11}{55}+\frac{10}{55}< \frac{x}{55}< \frac{22}{55}+\frac{1}{55}\)
\(\Rightarrow\frac{21}{55}< \frac{x}{55}< \frac{23}{55}\)
\(\Rightarrow21< x< 23\)
\(\Rightarrow x=22\)
\(2)\frac{11}{3}+\frac{-19}{6}+\frac{-15}{2}\le x\le\frac{19}{12}+\frac{-5}{4}+\frac{-10}{3}\)
\(\Rightarrow\frac{22}{6}+\frac{-19}{6}+\frac{-45}{6}\le x\le\frac{19}{12}+\frac{-15}{12}+\frac{-40}{12}\)
\(\Rightarrow\frac{22+\left[-19\right]+\left[-45\right]}{6}\le x\le\frac{19+\left[-15\right]+\left[-40\right]}{12}\)
\(=\frac{-42}{6}\le x\le\frac{-36}{12}\)
\(\Rightarrow-7\le x\le-3\)
\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)
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\(-2x-\left(x-17\right)=34-\left(-x+25\right)\)
\(-2x-x+17=34+x-25\)
\(-2x-x-x=34-25-17\)
\(-4x=-8\Leftrightarrow x=2\)
\(17x-\left(16x-37\right)=2x+43\)
\(17x-16x+37=2x+43\)
\(17x-16x-2x=-37+43\)
\(-x=6\Leftrightarrow x=6\)
\(-2x-3\left(x+17\right)=34-2\left(-x+25\right)\)
\(-2x-3x-51=34+2x-50\)
\(-2x-3x-2x=34-50+51\)
\(-7x=35\Leftrightarrow x=-5\)